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 Post subject: Geometry pg. 49 #19Posted: Wed Jul 27, 2011 6:59 pm

Joined: Mon Jun 20, 2011 9:51 pm
Posts: 5
Can you please help me understand this problem. Figure for this question would be very helpful.

Thanks.

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 Post subject: Re: Geometry pg. 49 #19Posted: Wed Jul 27, 2011 7:36 pm

Joined: Thu May 06, 2010 3:17 pm
Posts: 28
Location: Everywhere
The pyramid has four equal faces (which are triangles), and we're looking for surface area; essentially, we have to find the area of one of the triangles and multiply by four.

What do we need to find the area of the the triangle? The base and the height. We already have the length of the base. Since we know the square base of the whole pyramid is 36 square feet, each side must be 6 ft. B=6

What about height? The question gives us a height of 10, but that's for the entire pyramid. If we think about height, it's always measured vertically; for a pyramid, it wouldn't make sense to measure the height along the slant of one of the faces. The height is measured from the center of the base to the tip of the pyramid.

That height lets us make a right triangle with legs of the pyramid height (10) and half of the base (3, since we're measuring from the center of the base to one of the faces. The hypotenuse is the height of the triangular face.

a^2 + b^2 = c^2

10^2 + 3^2 = c^2

100 + 9 = c^2

109 = c^2

sqrt(109) = c

For the area of one of the faces:

A = (bh)/2

A = (6*sqrt(109))/2

A= 3*sqrt(109)

Multiply by 4 (since the pyramid has 4 faces):

4*3sqrt(109) = 12sqrt(109); answer B

I've attached a diagram of the whole pyramid and a cross-section. My iPad artistry isn't the greatest, so let me know if you need more help!

Bill

 Attachments: geometry19.png [ 68.19 KB | Viewed 1351 times ]
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 Post subject: Re: Geometry pg. 49 #19Posted: Wed Jul 27, 2011 8:09 pm

Joined: Mon Jun 20, 2011 9:51 pm
Posts: 5
Thanks Bill. Awesome and much simpler explanation.

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